In a certain obervation we got,`l = 23.2cm,r =1.32cm` and time taken for 10 oscillations was 10.0 s. Find, maximum percentage error in determinaton of 'g'.

Explanation

To find the maximum percentage error in the determination of 'g', we will follow these steps:

### Step 1: Understand the relationship between the variables
The time period \( T \) of a pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
From this, we can derive the expression for \( g \):
\[ g = \frac{4\pi^2 L}{T^2} \]

### Step 2: Identify the variables and their errors
Given:
- Length \( L = 23.2 \, \text{cm} \)
- Radius \( r = 1.32 \, \text{cm} \)
- Time for 10 oscillations \( T_{10} = 10.0 \, \text{s} \)

The effective length \( L \) (which includes the radius of the bob) is:
\[ L = 23.2 \, \text{cm} + 1.32 \, \text{cm} = 24.52 \, \text{cm} \]

### Step 3: Determine the errors in measurements
- The least count for measuring length is assumed to be \( 0.1 \, \text{cm} \), so the maximum error in \( L \) is:
\[ \Delta L = 0.1 \, \text{cm} \]
- The least count for measuring time is assumed to be \( 0.1 \, \text{s} \), so the maximum error in \( T \) for 10 oscillations is:
\[ \Delta T = 0.1 \, \text{s} \]
Thus, the time period \( T \) for one oscillation is:
\[ T = \frac{10.0 \, \text{s}}{10} = 1.0 \, \text{s} \]

### Step 4: Calculate the relative errors
The relative error in \( g \) can be expressed as:
\[ \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T} \]

Substituting the known values:
- \( L = 24.52 \, \text{cm} \)
- \( \Delta L = 0.1 \, \text{cm} \)
- \( T = 1.0 \, \text{s} \)
- \( \Delta T = 0.1 \, \text{s} \)

Calculating each term:
\[ \frac{\Delta L}{L} = \frac{0.1}{24.52} \approx 0.00407 \]
\[ 2 \frac{\Delta T}{T} = 2 \times \frac{0.1}{1.0} = 0.2 \]

Now, adding these relative errors:
\[ \frac{\Delta g}{g} \approx 0.00407 + 0.2 = 0.20407 \]

### Step 5: Convert to percentage error
To find the percentage error, multiply by 100%:
\[ \text{Percentage error} = \frac{\Delta g}{g} \times 100\% \approx 0.20407 \times 100\% \approx 20.41\% \]

### Final Answer
The maximum percentage error in the determination of \( g \) is approximately \( 20.41\% \).

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